Question

A simple pendulum consists of a 4.9-kg mass attached to a string. The pendulum is pulled to the right and held at rest so that it is 2.08 m above the lowest point in its swing. If you let it go from this point, how fast (in m/s) is it traveling at the lowest point of its swing (on the initial pass)

Answer 1

Answer:

6.384 m/$sec^{2}$

Explanation:

The velocity is given by;

We know;

PE= mgh

KE= 1/2 m$V^{2}$

but KE=PE

==> 1/2 m$V^{2}$ =mgh

==> 1/2 $V^{2}$ =gh

==> $v=\sqrt{2gh}$

puting g=9.8 m/sec2

h=2.08 m

==> $v= 6.384 m/sec^{2}$=